### High and Low pass Fillter

```Lecture 23
Filters
Hung-yi Lee
Filter Types
Lowpass filter
Bandpass filter
wco : cutoff frequency
Bandwidth B = wu - wl
Highpass filter
Notch filter
Real World
Ideal filter
Transfer Function – Rules
• Filter is characterized by its transfer function
H s  =
N s 
D s 
=K
 s  z1  s  z 2   s  z m 
 s  p1  s  p 2   s  p n 
The poles should be at the
left half of the s-plane.
We only consider
stable filter.
Given a complex pole or zero, its complex
conjugate is also pole or zero.
Transfer Function – Rules
• Filter is characterized by its transfer function
H s  =
N s 
D s 
=K
 s  z1  s  z 2   s  z m 
 s  p1  s  p 2   s  p n 
n  m :improper filter
As the frequency increase,
the output will become
infinity.
n  m :proper filter
We only consider proper
filer.
The filters consider have more poles than zeros.
Filter Order
H s  =
N s 
D s 
=K
 s  z1  s  z 2   s  z m 
 s  p1  s  p 2   s  p n 
Order = n
The order of the denominator is the order of the filter.
order=1
order=4
order=4
Outline
• Textbook: Chapter 11.2
Second-order Filter
First-order Filters
Lowpass
Filter
Highpass
Filter
Lowpass
Filter
Highpass
Filter
Bandpss
Filter
Notch
Filter
First-order Filters
Firsr-order Filters
H s  =
N s 
zero or first order
first order
D s 
0 or 1 zero
1 pole
Case 1:
Case 2:
1 pole, 0 zero
1 pole, 1 zero
H s  = K
1
s p
H s  = K
sz
s p
Firsr-order Filters - Case 1
Lowpass filter
H s  = K
w
1
s p
As ω increases
Magnitude decrease
Phase decrease

Pole p is on the
negative real axis
Firsr-order Filters - Case 1
• Amplitude of the transfer function of the first-order
low pass filter
Ideal
Lowpass
filter
First-order
Lowpass
filter
Firsr-order Filters - Case 1
• Find cut-off frequency ωco of the first-order low
pass filter
w
At DC
Lowpass filter
H s  = K
 0  = K
1
s p
1
| p|
Find cut-off frequency ωco
such that
| p|

 w co  = K
1
| p|
1
2
w co = | p |
Firsr-order Filters - Case 2
H s  = K
sz
s p
w
Zero can be
positive or
negative
|z|
| p|
Case 2-1: Absolute value of zero is
smaller than pole
 Magnitude is proportional to the
length of green line divided by the
length of the blue line
 Low frequency ≈ |z|/|p|
Because |z|<|p|
The low frequency signal will
be attenuated
If z=0, the low frequency can

be completely block
Not a low pass
Firsr-order Filters - Case 2
H s  = K
sz
s p
w
Case 2-1: Absolute value of zero is
smaller than pole
 Magnitude is proportional to the
length of green line divided by the
length of the blue line
 High frequency
The high frequency signal
will pass
|z|
| p|

If z=0 (completely block low
frequency)
H s  = K
s
s p
High pass
First-order Filters - Case 2
H s  = K
s
s p
• Find cut-off frequency ωco of the first-order high
pass filter
 w  = K
| jw |
w co
| jw  p |
w |p|
2
2
co
   = K
 w co  = K
2
=
1
2
w co = | p |
| j w co |
| j w co - p |
=
K
2
(the same as low
pass filter)
First-order Filters - Case 2
H s  = K
sz
s p
w
Case 2-2: Absolute value of zero is
larger than pole
 Low frequency ≈ |z|/|p|
Because |z|>|p|
The low frequency signal will
be enhanced.
 High frequency: magnitude is 1

The high frequency signal
will pass.
Neither high pass
nor low pass
First-order Filters

vh
1
H lp  s  =

1
sC
R
=
1
1
sRC  1
sC
1
vl
1
H lp  j w  =
 w  =
RC
jw 
1
Lowpass filter
If vh is output
Highpass filter
 0  = 1
   = 0
RC
 1 
2
w 

RC


RC
If vl is output
RC
1
s
RC
1
sC
Consider vin as input
=
w co =
1
2
(pole)
RC
s
H hp  s  = 1  H lp  s  =
s
1
RC
 0  = 0
   = 1
w co =
1
RC
(pole)
First-order Filters
H lp  s  =

vh
R
R
sL  R
=
L
s

R
L
R
R
H lp  j w  =
vl
L
jw 
R
 w  =
L
R 
w  
L
2
L
 0  = 1    = 0 w co =
H hp  s  = 1  H lp  s  =
2
s
s
R
L
R
L
(pole)

V out
Vx
V in

H lp 1  s  =
H lp  s  =
Vx
V in
H lp 2  s  =
H lp 1  s 

V out
Vx
H lp 2  s 

V out
Vx
V in

1
1
H lp  s  =
=
sC 1
R1 
1
sC 1

sC 2
R2 
=
1
1
sC 1 R1  1
sC 2
1
1  s C 1 R 1  C 2 R 2   s C 1 C 2 R 1 R 2
2

1
sC 2 R 2  1

V out
Vx
V in

Z eq  s 
The first low pass filter is influenced by the
second low pass filter!
H lp 1  s  =
Vx
V in
=
Z eq  s 
Z eq  s   R1

V out
Vx
V in


1 


Z eq  s  =
||  R 2 
sC 1 
sC 2 
1 
1 
 R2 


sC 1 
sC 2  
=

1
1  

  R 2 
sC 1 
sC 2 
Z eq  s 
1
2
s C 1C 2
2
s C 1C 2
=
sC 2 R 2  1
s C 1  C 2   s C 1 C 2 R 2
2

V out
Vx
V in

Z eq  s 
H lp  s  = H lp 1  s   H lp 2  s  =
=
1
Z eq  s 
Z eq  s   R1
sC 2
1
sC 2
1
1  s  R 1C 1  C 2 R 2  R 1C 2   s C 1C 2 R 1 R 2
2
 R2
Second-order Filters
Second-order Filter
H s  =
N s 
0, 1 or 2 zeros
D s 
Second order
Must having
two poles
2 poles
Case 1:
No zeros
Case 2:
One zeros
Case 3:
Two zeros
Second-order Filter – Case 1
Case 1-1
Case 1-2
w
w

p1
p1
p2
p2

Second-order Filter – Case 1
Case 1-1
Real Poles
w
The magnitude is
l1
The magnitude
monotonically decreases.
l1
p1
p2
l1l 2
As ω increases
l2
l2
K

Decrease faster than first
order low pass
Second-order Filter – Case 1
Case 1-2
Complex Poles
w
The magnitude is
l1
As ω increases,
l2
p1
l1
l2
p2
K
l1l 2
l1 decrease first and then increase.

l2 always increase
What will happen to magnitude?
1. Increase
2. Decrease
3. Increase, then decrease
4. Decrease, then increase
Second-order Filter – Case 1
Case 1-2
Complex Poles
w
If ω > ωd
l1
l2
p1
wd
wd
p2
l1 and l2 both increase.

The magnitude must decrease.

What will happen to magnitude?
1. Increase
2. Decrease
3. Increase, then decrease
4. Decrease, then increase
Second-order Filter – Case 1
Case 1-2
Complex Poles
l1
p1
wd
wd
p2

When ω < ωd
w
w

l2
l1 =
  w d  w 
l2 =
  w d  w 
2
2
2
2
Maximize the magnitude
K
l1l 2
Minimize
l1l 2 =

2

 wd  w

2

2

 wd  w

2

Second-order Filter – Case 1
Minimize l l =   w  w    w  w  
Minimize f w  =   w  w    w  w  
df w 
= 2 w  w   1   w  w    2 w  w   w
dw
w  w  1  w  w    w  w   w  w   = 0
2
2
1 2
2
2
d
d
2
2
2
2
d
d
2
2
d
2

2
 2 w   w d
  w d  w
2

2

w
d
= 0
 
 w  wd  w


2
= wd  w
2


=0
d
 w w d  w  2w  = 0
2
w
2
2
d
 w wd  w
d
2
d
 2 w   w d
d
2
d

2
d
2
w  wd
2
 = 0
  w d  w w d  w  = 0
2
w =
wd  
2
2
(maximize)
Second-order Filter – Case 1
w =
wd  
2
2
The maxima exists when w d  
w
Peaking
w
p1
p1
wd
wd

wd
wd
p2
  wd


p2
No Peaking
wd  
Peaking

Second-order Filter – Case 1
w =
wd  
2
2
The maxima exists when w d  
wd
l1
w

wd
p2
 w  =
w
p1
=K

l2
Peaking
K
l1l 2
1
  w d  w 
2
2
  w d  w 
2
2
Assume w = w d2   2
 w  = K
1
2w
 0  = K
d
1
  wd
2
2
Second-order Filter – Case 1
K
H s  =
s 
2
w0
Q
sw
2
0

w0

Q
p1 , p 2 =
 w0

 Q
2

  4 w 02

2
For complex poles
p1
w0
wd
wd


  4 w 02  0


 =
w0
  wd = w0
2
Q 
wd = w0 1
2Q
w0
p2
 w0

 Q

2
2
1
2
1
4Q
2
Second-order Filter – Case 1
H s  =
s 
2
w0
Q
sw
 jw 
2
0
w = w0
=
Q times
p1
K
H  jw  =
K
w
w0
wd
wd
w0

Q
K
w
2

jw  w 0
2
j
w0
w
Q
K
H  jw 0  =
w = w0
w0
2
j

w0
p2
2
0
2
Q
 w 0  =
K
w
2
0
Q
 0  =
K
w0
2
Q times of DC gain
Second-order Filter – Case 1
w =
w 
2
d
2
K
H s  =
s 
2
w0
 w  = K

sw
Q
For complex poles
2
0
 w0

 Q

 =
p1 , p 2 =
w0
Q

 w0

 Q
2
2
w0
2Q
wd = w0 1
2w

  4 w 02

2

  4 w 02  0


1
1
Q 
2
1
4Q
2
d
Second-order Filter – Case 1
w =
w 
2
d
2
H s  =
s 
2
w0
 =
s  w0
2
Q
w = w0 1
 w  = K
K
1
2Q
2
w0
wd = w0 1
2Q
1
2w
d
1
4Q
2
The maximum value is K
The maximum exist when Q 
1
2
1
2w
= 0 . 707
= K
d
1
Q
w0
2
1
1
4Q
2
Second-order Filter – Case 1
Case 1-1
Real Poles
Case 1-2
Complex Poles
w
w
p1
p1
p2

wd
wd
Q  0 .5

(No Peaking)
p2
0 . 707  Q  0 . 5
Which one is considered as closer
to ideal low pass filter?

K
H s  =
s 
2
w0
Q
Q  0 .5
s  w0
2
Q =
1
Complex
poles
= 0 . 707
2 (Butterworth filter)
Q  0 . 707
Peaking
Butterworth – Cut-off Frequency
K
H s  =
s 
2
w0
s  w0
H  jw  =
=
 0  =
K
s 
2
= 0 . 707
2
2
Q
=
Q =
1
2w 0 s  w 0
2
K
w0
2
 w co  =
K
 j w 2 
w
2w 0  jw   w 0
2
K
2
0
w
2

j 2w 0 w
1
K
2 w0
2
w co = w 0
ω0 is the cut-off frequency
for the second-order
lowpass butterworth filter
(Go to the next lecture first)
Second-order Filter – Case 2
Case 2: 2 poles and 1 zero
Case 2-1: 2 real poles and 1 zero
w
z1
p 2 p1
w

p2
z1 p1

w
p 2 p1
z1

Second-order Filter – Case 2
Case 2: 2 poles and 1 zero
Case 2-1: 2 real poles and 1 zero
w
p 2 p1
z1
flat

Bandpass Filter
Second-order Filter – Case 2
Case 2-2: 2 complex poles and 1 zero
w
Two
Complex
Poles
p1
-40dB
w0
w0
+

z1
Zero
+20dB
p2
| z1 |
Second-order Filter – Case 2
Case 2-2: 2 complex poles and 1 zero
| z 1 | w 0
-40dB
-20dB
Two
Complex
Poles
w 0 | z1 |
-40dB
w0
+
+20dB
-20dB
Zero
+20dB
| z 1 | w 0
| z1 |
w0
| z1 |
Second-order Filter – Case 2
Case 2-2: 2 complex poles and 1 zero
Highly Selective
+20dB
-20dB
| z 1 |= 0
Two
Complex
Poles
-40dB
w0
+
w0
Zero
+20dB
Bandpass Filter
| z1 |
Bandpass Filter
• Bandpass filter: 2 poles and zero at original point
H bp ( s ) = K
 w 
s
s  w 0 Q s  w 0
2
2
Find the frequency for the
maximum amplitude
bandpass filter
ω0?
Bandpass Filter
• Find the frequency for the maximum amplitude
H bp ( s ) = K
H bp ( j w ) =
=
s
s  w 0 Q s  w
2
2
0
K  w 0 Q  j w
 jw 
2
 w 0 Q  j w  w
K
jw Q
w0
1
w0Q
jw
=
2
0
=
=
w 0
K 2
s  w 0
w0
Q s  w 0
2
K
 j w 2
w 0 Q  j w
K
jw Q
Q s
1
jw 0 Q
w
=
w0
2
1
w 0
Q  jw
K
 w
w0 


1  jQ 

w 
 w0
Bandpass Filter
• Find the frequency for the maximum amplitude
H bp ( j w ) =
K
 w
w0 

1  jQ 


w
w
 0

K
a bp (w ) =
 w
w0 

1  Q 


w
w
 0

2
a bp (w ) is maximized
when w = w 0
(Center frequency)
The maximum value is K’.
2
a bp ( 0 ) = 0
a bp (  ) = 0
(Bandpass filter)
Bandpass Filter
a bp (w ) is maximized
K
a bp (w ) =
w0 
2 w

1  Q 


w
w
 0

 w 
when w = w 0
2
The maximum value is K’.
bandpass filter
K
K / 2
B
wl
w0
Bandwidth B
= ωr - ωl
wr
Bandpass Filter - Bandwidth B
K
a bp (w ) =
=
w0 
2 w

1  Q 


w
w
 0

2
K
2
2
2
 w
w0 
1


w  w  = Q2
 0

1
2
2
w  w 0w  w 0 = 0
Q
 w
w0 
 =2
1  Q 


w
w
 0

w0
w
1

=
w0 w
Q
2
2

w =
1
Q
w0 
 1

2
 w 0   4 w 0
Q

2
Pick the two positive
ones as ωl or ωr
Bandpass Filter - Bandwidth B
2

w =
1
w0 
Q
 1

2
 w 0   4 w 0
Q

2
2
1
wr =
Q
w0 
 1

2
 w 0   4 w 0
Q

2
B = wr  wl =
w0
Q
w = w rw l
2
0
Q measure the narrowness
of the pass band
Q is called quality factor
2

wl =
1
Q
w0 
 1

2
 w 0   4 w 0
Q

2
H bp ( s ) = K
= K
s
s  w 0 Q s  w 0
2
2
s
s  Bs  w0
2
2
Bandpass Filter
H bp ( s ) = K
= K
s
s
2
 w 0
2

Q s  w0
s
s  Bs  w0
2
2
 Usually require a specific bandwidth
 The value of Q determines the
bandwidth.
 When Q is small, the
transition would not be
sharp.
Stagger-tuned Bandpass Filter
Stagger-tuned Bandpass Filter
- Exercise 11.64
s
H s  = K
s 
2
w1
K = 1600
s
sw
Q
Bandpass Filter
Center frequency: 10Hz
2
1
s 
2
w2
sw
Q
w 1 = 10
w 2 = 40
2
2
Bandpass Filter
Center frequency: 40Hz
We want flat passband.
dB
Tune the value of
Q to achieve that
10 Hz
40 Hz
Stagger-tuned Bandpass Filter
- Exercise 11.64
H s  = K
s 
2
s
s
w1
w2
Q
s  w1 s 
2
2
Q
K = 1600
s  w2
2
w 1 = 10
w 2 = 40
Test Different Q
Q=3
Q=1
Q=0.5
Second-order Filter – Case 3
• Case 3: Two poles, Two zeros
Case 3-1: Two real zeros
Two real
poles
Two
Complex
poles
w
w
p1
z1
z1
p2
p1
z2

z2
p2
High-pass

Second-order Filter – Case 3
Case 3: Two poles, Two zeros
Case 3-2: Two Complex zeros
s 
p 1  s  p 2  = s 
2
w0
Q
w0
p1
w
w
s  w0
2
z1
Fix ω0
Larger Q
 s  z1  s  z 2  =

Larger θ
s 
2
w
Q

s  w
2
z2
Fix ωβ
Larger Q β

Larger θ β
p2
Second-order Filter – Case 3
Case 3: Two poles, Two zeros
Case 3-2: Two Complex zeros
w0  w
w0
p1
w
w
z1
Two
poles
w0
-40dB



Two
zeros
z2
w
+40dB
p2
Second-order Filter – Case 3
Case 3: Two poles, Two zeros
Case 3-2: Two Complex zeros
w0  w
s 
2
w
Q
H s  =
s 
2
w0
Q
w0
p1
w
w
s  w
2
z1
s  w0
2

w 0 = 10, w  = 1, Q = Q  = 10


z2
High-pass
Notch
p2
Second-order Filter – Case 3
Case 3: Two poles, Two zeros
Case 3-2: Two Complex zeros
w
w0  w
w0
w
z1
p1
Two
poles
w0
-40dB



Two
zeros
w
+40dB
p2
z2
Second-order Filter – Case 3
Case 3: Two poles, Two zeros
Case 3-2: Two Complex zeros
w0  w
s 
2
w
Q
H s  =
s 
2
w0
Q
s  w
w
w
w0
2
z1
p1
s  w0
2

w 0 = 1, w  = 10, Q = Q  = 10
Low-pass
Notch


p2
z2
Second-order Filter – Case 3
Case 3: Two poles, Two zeros
Case 3-2: Two Complex zeros
w0 = w
  
Q  Q
w0 = w
w
p1
z1
Large Q
Two
poles
Two
zeros
w0

-40dB


z2
w
+40dB
small Qβ
p2
Second-order Filter – Case 3
Case 3: Two poles, Two zeros
Case 3-2: Two Complex zeros
w0 = w
  
Q  Q
w0 = w
w
p1
z1
w 0 = w  = 10
Q = 100, Q  = 10



z2
p2
Second-order Filter – Case 3
Case 3: Two poles, Two zeros
Case 3-2: Two Complex zeros
w0 = w
Two
poles
Two
zeros
  
Q  Q
w0 = w
w
p1
z1
small Q
w0
w

-40dB


+40dB
Larger Qβ
p2
z2
Second-order Filter – Case 3
Case 3: Two poles, Two zeros
Case 3-2: Two Complex zeros
w0 = w
  
Q  Q
w0 = w
w
p1
z1
w 0 = w  = 10
Q = 10, Q  = 100
Standard
Notch
Filter

p2


z2
Second-order Filter – Case 3
Case 3: Two poles, Two zeros
Case 3-2: Two Complex zeros
w0 = w
s 
w0
Q
H s  =
s 
2
w0
Q
w 0 = w  z1
Q  Q
  
If the two zeros
are on the ω axis
The notch filter will
completely block the
frequency ω0
2
w
p1
Q = 

s  w0


2
s  w0
2
=
sw
2
0
s 
2
w0
Q
2
sw
2
0
p2
z2
Notch Filter
s 
2
w0
Q
H no  s  =
s 
2
w0
Q
w
 no w  =
w
2
0
2
0
s  w0
2
The extreme value
is at ω= ω0
s  w0
2
 no w 0  =

w w
 0
 Q
 




2

 w 0w
 
 Q



2
w
2 2
w
2 2
 no 0  = 1
Q
Q
 no   = 1
(Notch filter)
Second-order RLC Filters
A

v in
C
v in
v out


v out

B
v out

v in

D
v in

v out

RLC series circuit can implement high-pass, lowpass, band-pass and notch filter.
Second-order RLC Filters
A

v in
v out

DC (O)
Infinity (X)
Low-pass Filter

B
v out

v in
DC (X)
Infinity (O)
High-pass Filter
Second-order RLC Filters

v out
w

C
p1
v in

z1
p2
H s  =
R
R  sL 
1
=
sCR
sC R  s CL  1
2
s
=
s s
2
R
L
R
L
sC
Band-pass Filter

1
LC
Second-order RLC Filters
– Band-pass
C

v out

v in
40pF to 360pF
L=240μH, R=12Ω
s
H s  =
s s
2
R
L
R
L
w0
Q
Frequency range
Center frequency: w 0 = 1

1
LC
w
2
0
f0 =
Max: 1.6MHz
w0
2
LC
= 1 2
LC
min: 0.54MHz
Second-order RLC Filters
– Band-pass
C

v out

v in
40pF to 360pF
L=240μH, R=12Ω
s
H s  =
s s
2
Frequency range 0.54MHz ~ 1.6MHz
R
L
R
L
w0
Q
w0 = 1

1
LC
w
2
0
Q =
LC
w 0L
R
w0 Q = R L
Q =
1
L
R
C
Q is 68 to 204.
Band-pass
1
s
v in
1
s
Z eq
v out
s
1
s

s 1
s
=
s 1
s s  2 
s
V out = V in

1
1  s 1

= ||  1   = || 

s 
s
s  s 
1
1 s 1
=

Z eq
1  Z eq
s 1
1
1
1
s
s
s s  2 
= V in
s 1 s 1
1
s s  2 
= V in
s
s  3s  1
2
Band-pass
1
s
v in

1
v out
s
H s  =
s
s  3s  1
2

Band-pass Filter
V out = V in
H s  =
V out
V in
=
s
s  3s  1
2
s
s  3s  1
2
w =1
2
0
w0 = 1
B=
w0
Q
Q =
1
3
=3
Second-order RLC Filters
C

v out

D

v out
v in
v in

s
H bp  s  =
s s
2
H  s  = 1 - H bp  s 
R
L
R
L

LC
1
s 
2
1
LC
=
s s
2
R
L
Notch Filter

1
LC
Active Filter
V in = i  Z i
Basic Active Filter
V out =  i  Z f
-i
H s  =
i
0
0
V in
V out
V out
V in
=
Z
f
Zi
First-order Low-pass Filter
Z
R
f
H s  = 
f
=
V out
=
||
f
sC
=
Ri
Zi
R
Zi
V in
Z
1

R iR


1
f
sC
f
R
f
1 


sC f 
f
R i 1  sR f C f
w co = 1 R f C f

f
First-order High-pass Filter
Z
H s  = 
f
Zi
=
Z
f
Zi
R
Ri 
V in
V out
=
f
1
sC i
sC i R
f
sC i R i  1
w co = 1 R i C i
Active Band-pass Filter
Band-pass Filter
Active Band-pass Filter
H 1 s  = 
=
R
R i 1  sR f C f
R
1  sR C 1
H 2 s  = 
=

R
H bp s  = H 1 s H 2 s  =  
 1  sRC 1
f
sC i R
f
sC i R i  1
sR C 2
sR C 2  1

sR C 2 
 

  sR C  1 
2


?

H s  = 
v out
Z
the transfer function of
passive filters.
v in
Z
f
Zi
v out
change the transfer
function of the active filter.
Z
H s  =
The transfer function is H(s).
V in
H  s V in
One Filter Stage Model
V out
V in
V out
V1
H 1  s V1
1st Filter with
transfer function H1(s)
H 2  s V 2
V2
2st Filter with
transfer function H2(s)
Overall Transfer Function: H  s  =
V3
V1
 H 1  s H 2  s 
V3
H 1  s V1
V1
1st Filter with
transfer function H1(s)
V 2 = H 1  s V1
V 3 = H 2  s V 2
Z i2
Z o1  Z i 2
H 2  s V 2
V2
V3
2st Filter with
transfer function H1(s)
V 3 = H 2  s H 1  s V1
H s  =
V3
V1
Z i2
Z o1  Z i 2
= H 2  s H 1  s 
Z i2
Z o1  Z i 2
H  s  = H 2  s H 1  s 
V1
H 1  s V1
1st Filter with
transfer function H1(s)
Z i2
Z o1  Z i 2
H 2  s V 2
V2
2st Filter with
transfer function H1(s)
H  s  = H 2  s H 1  s 
If zero output impedance (Zo1=0)
or If infinite input impedance (Zi2=∞)
V3
– Input & Output Impedance
Z o s 
IT
VT
Z i s 
Z i s  =
H  s V i
VT
IT
Z o s 
Vi = 0
Z i s 
H  s V i V T
IT
Z o s  =
VT
IT
H  s  = H 2  s H 1  s 
If zero output
impedance (Zo1=0)
or If infinite input
impedance (Zi2=∞)
– Basic Active Filter
-i =0
i =0
Z o s  =
0
0
0
V in
=0
IT
VT
VT
IT
=0
Active Notch Filter
A
B
Which one is correct?
Active Notch Filter
Low-pass
Filter
High-pass
Filter
Together
Homework
• 11.19
Thank you!
• 11.19: Ra=7.96kΩ, Rb= 796Ω,
va(t)=8.57cos(0.6ω1t-31。)
+0.83cos(1.2ω2t-85。)
vb(t)=0.60cos(0.6ω1t+87。)
+7.86cos(1.2ω2t+40。)
(ω1 and ω2 are 2πf1 and 2πf2 respectively)
• 11.22: x=0.14, ωco=0.374/RC
• 11.26(refer to P494): ω0=2π X 6 X 10^4, B= ω0=2π X 5 X
10^4, Q=1.2, R=45.2Ω, C=70.4nF
• 11.28(refer to P494): C=0.25μF, Qpar=100, Rpar=4kΩ,
R||Rpar=2kΩ, R=4kΩ
Acknowledgement
• 感謝 江貫榮(b02)
• 上課時指出投影片的錯誤
• 感謝 徐瑞陽(b02)
• 上課時糾正老師板書的錯誤
Appendix
Aliasing
Sampling
Wrong
Interpolation
Actual
signal
High frequency becomes low frequency
Phase
cos 10 t 
 cos  20 t 
filter  w  = 1  w  = 90 
 

cos  10 t  
2

 

 cos  20 t  
2

Table 11.3 Simple Filter
Type
Lowpass
Highpass
Bandpass
Notch
Transfer Function
H (s) =
H (s) =
H (s) =
H (s) =
Properties
a (0 ) = K
K w co
s  w co
a (w co ) = K
a ( ) = K
Ks
s  w co
a (w co ) = K
K w 0 Q s
2
K s  2s  w0
2
2
2
B = w0 Q

a (w 0 ) = KQ  w 0
s  w 0 Q s  w 0
2
2
a (w 0 ) = K
s  w 0 Q s  w 0

2
2
B = w0 Q
98
Loudspeaker for home usage with three
types of dynamic drivers
1. Mid-range driver
2. Tweeter
3. Woofers
From Wiki
• Butterworth filter – maximally flat in passband and
stopband for the given order
• Chebyshev filter (Type I) – maximally flat in stopband,
sharper cutoff than Butterworth of same order
• Chebyshev filter (Type II) – maximally flat in passband,
sharper cutoff than Butterworth of same order
• Bessel filter – best pulse response for a given order
because it has no group delay ripple
• Elliptic filter – sharpest cutoff (narrowest transition
between pass band and stop band) for the given order
• Gaussian filter – minimum group delay; gives no
overshoot to a step function.
• http://www.ti.com/lsds/ti/analog/webench/weben
ch-filters.page
• http://www.analog.com/designtools/en/filterwizar
d/#/type
Suppose this band-stop filter were to
suddenly start acting as a high-pass
filter. Identify a single component
failure that could cause this problem
to occur:
If resistor R3 failed open, it would cause
this problem. However, this is not
the only failure that could cause the same
type of problem!
```